Proof for $\nabla \cdot \iiint A dv$=$\iiint \nabla \cdot A dv$

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if $V=\iiint A dv$ Where $A$ and $V$ are vectors Then taking the divergence of both sides

$\nabla \cdot V$ = $\nabla \cdot \iiint A dv$=$\iiint \nabla \cdot A dv$

My mathematical knowledge doesn't exceeds calculus 3 and I found this result in a physics book.

Intuitively it's true but I'm asking for a rigorous proof of it.